Based on `VSEPR` theory the number of `90^(@) F -Br -F` angles in a molecules of `BrF_(5)` is_____.

Explanation

To determine the number of 90° F-Br-F angles in the molecule BrF₅ based on VSEPR theory, we can follow these steps:

### Step 1: Determine the Valence Electrons
- Bromine (Br) has 7 valence electrons.
- Each fluorine (F) atom has 7 valence electrons, and since there are 5 fluorine atoms, they contribute a total of 35 valence electrons (5 × 7 = 35).
- Therefore, the total number of valence electrons in BrF₅ is 7 (from Br) + 35 (from 5 F) = 42 valence electrons.

### Step 2: Bonding and Lone Pairs
- In BrF₅, bromine forms 5 bonds with the 5 fluorine atoms, using 5 of its valence electrons.
- This leaves 2 electrons from bromine, which will be present as a lone pair.

### Step 3: Determine the Molecular Geometry
- According to VSEPR theory, the presence of 5 bonding pairs and 1 lone pair around the central bromine atom leads to a square pyramidal geometry.
- The square pyramidal shape consists of a square base formed by 4 fluorine atoms and one fluorine atom at the apex.

### Step 4: Analyze the Bond Angles
- In a perfect square pyramidal structure, the angles between the equatorial fluorine atoms (the ones in the square base) would ideally be 90°.
- However, the presence of the lone pair causes repulsion, which affects the bond angles.

### Step 5: Conclusion on 90° Angles
- Due to the repulsion between the lone pair and the bonding pairs, the bond angles are not exactly 90°; they are approximately 84.8°.
- Therefore, there are no angles that are exactly 90° in the BrF₅ molecule.

### Final Answer
The number of 90° F-Br-F angles in BrF₅ is **0**.

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